Question

If ${\sec }^2\theta +{\tan }^2\theta +1=2$, then find the value of $\sec (-\theta )$:

• A. $-2$
• B. $-\frac{1}{2}$
• C. $1$
• D. $\pm 1$

Answer 1

The correct option is D $\pm 1$

Given : ${\sec }^2\theta +{\tan }^2\theta +1=2$

$⟹$ $\frac{1}{{\cos }^2\theta }+\frac{{\sin }^2\theta }{{\cos }^2\theta }+1=2$

$⟹$ $1+{\sin }^2\theta +co{s}^2\theta =2{\cos }^2\theta$

$⟹$ $1+1=2co{s}^2\theta$

$⟹$ $co{s}^2\theta =1$

$⟹$ $\cos \theta =\pm 1$

$\because$ $\cos (-\theta )=\cos \theta$

$\therefore$ $\sec (-\theta )=\sec \left(\theta \right)=\frac{1}{\cos \theta }=\pm 1$